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\hypersetup{
  pdftitle={Ch3 人寿保险},
  pdfauthor={滕帆},
  hidelinks,
  pdfcreator={LaTeX via pandoc}}

\title{Ch3 人寿保险}
\subtitle{保险精算学}
\author{滕帆}
\date{2023-09}
\institute{浙大宁波理工学院金融学系}

\begin{document}
\frame{\titlepage}

\begin{frame}[allowframebreaks]
  \tableofcontents[hideallsubsections]
\end{frame}
\hypertarget{ux5982ux4f55ux7528ux6570ux5b66ux6765ux63cfux8ff0ux4ebaux5bffux4fddux9669}{%
\section{\texorpdfstring{如何用\textbf{数学}来描述人寿保险？}{如何用数学来描述人寿保险？}}\label{ux5982ux4f55ux7528ux6570ux5b66ux6765ux63cfux8ff0ux4ebaux5bffux4fddux9669}}

\begin{frame}{如何用\textbf{数学}来描述人寿保险？}
\begin{itemize}
\item
  人寿保险：以被保险人的身故作为给付条件的保险。传统人寿保险主要分为定期寿险、终身寿险和两全保险。
\item
  ``\textbf{事后诸葛亮}''案例
\end{itemize}

\begin{quote}
假设有1000位(45)男性群体投保了一年期定期寿险，未来一年内\textbf{一定}会有12位被保险人身故，合同约定身故保险金为1万元。暂不考虑\textbf{利率}，请问保险公司在一年前至少收多少钱的保费？
\end{quote}

\begin{itemize}
\tightlist
\item
  当我们把''\textbf{一定}''去掉，且考虑''\textbf{利率}''，保险公司的给付金额与身故人数（死亡率）、身故时间密切相关。
\end{itemize}
\end{frame}

\begin{frame}
\begin{columns}[T]
\begin{column}{0.45\textwidth}
\begin{quote}
假设有1000位(45)男性群体投保了一年期定期寿险，未来一年内\textbf{一定}会有12位被保险人身故，合同约定身故保险金为1万元。请问保险公司在一年前至少收多少钱的保费？
\end{quote}
\end{column}

\begin{column}{0.55\textwidth}
假定：

\begin{enumerate}
\item
  身故人数不变且均匀分布，即每个月1人身故。
\item
  身故月末给付1万元。
\item
  年实际收益率3\%。
\end{enumerate}

如何计算给付金额。
\end{column}
\end{columns}
\end{frame}

\begin{frame}
真实世界里的人寿保险

\begin{itemize}
\item
  期限通常超过1年，大部分超过10年
\item
  每个年龄的死亡率不尽相同，且身故人数不一定均匀分布
\item
  即使死亡率保持稳定，但身故人数仍然不确定，同样需要考虑风险
\end{itemize}
\end{frame}

\begin{frame}
接下来，我们将通过由浅入深的方式来介绍寿险精算的基本内容。
\end{frame}

\hypertarget{ux5b9aux671fux5bffux9669}{%
\section{定期寿险}\label{ux5b9aux671fux5bffux9669}}

\begin{frame}{定期寿险}
假设1（保险期限）：投保年龄为x，保险期限为n年

假设2（预定利率）：年实际利率为i

假设3（趸交保费与精算现值）：保费在投保时一次性缴纳。\textbf{寿险精算现值}是指在投保时的保险赔付期望现值。简言之，精算现值即纯保费，是毛保费的重要组成部分。

假设4（身故保险金与赔付时间）：被保险人身故后固定给付1元，赔付时间简单设定为\textbf{身故年末}/\textbf{身故月末}/\textbf{即死即付}三种模式。
\end{frame}

\begin{frame}{身故年末给付的定期寿险精算现值}
\protect\hypertarget{ux8eabux6545ux5e74ux672bux7ed9ux4ed8ux7684ux5b9aux671fux5bffux9669ux7cbeux7b97ux73b0ux503c}{}
定义''身故年末给付的定期寿险精算现值''的精算符号为

\[
\Ax{\termxn}
\]

其中：A为''Assurance''(寿险)的简写，A右下角的x表示投保年龄，n为保险期限。A右上角左侧的''1''表示因身故给付。因为年实际利率预定为i，暂忽略不写。
\end{frame}

\begin{frame}{\(\Ax{\termxn}\)计算公式}
\protect\hypertarget{axtermxnux8ba1ux7b97ux516cux5f0f}{}
\begin{itemize}
\item
  按照定期寿险精算现值的定义，保险公司向\(\lx{x}\)收取每个人\(\Ax{\termxn}\)的趸交纯保费。
\item
  在后面的n年内，每年末身故人数为\(\dx{x+t}\)，其中\(0\leq t \leq n-1\)
\item
  显然
\end{itemize}

\[\lx{x} \times \Ax{\termxn} = \dx{x} v+\dx{x+1} v^2+ \ldots + \dx{x+n-1} v^n\]

\begin{itemize}
\tightlist
\item
  将上式两边同除以\(\lx{x}\)
  \footnote{$\qx[s|]{x}=\px[s]{x}\times\qx{x+s}$}
\end{itemize}

\[\Ax{\termxn} = \qx{x} v+\qx[1|]{x} v^2+ \ldots + \qx[n-1|]{x} v^n\]
\end{frame}

\begin{frame}
利用\textbf{生命表}计算\(\Ax{\termxn}\)

\begin{itemize}
\item
  观察\(\lx{x} \times \Ax{\termxn} = \dx{x} v+\dx{x+1} v^2+ \ldots + \dx{x+n-1} v^n\)
\item
  将上式两边同乘以\(v^x\)变作
\end{itemize}

\[
\begin{aligned}
\lx{x}v^x \times \Ax{\termxn} \\
&= \dx{x} v\times v^x+\dx{x+1} v^2\times v^x+ \ldots + \dx{x+n-1} v^n\times v^x \\
&= \dx{x} v^{x+1} +\dx{x+1} v^{x+2} +\ldots +\dx{x+n-1} v^{n+x} \\
&= \sum_{t=0}^{n-1} \dx{x+t}v^{x+t+1}
\end{aligned}
\]

令\(C_x=\dx{x}v^{x+1}\)、\(D_x=\lx{x}v^x\)，那么

\[
\Ax{\termxn} = \frac{\sum_{t=0}^{n-1} C_{x+t}}{D_x}
\]

\begin{block}{Excel应用举例}
\protect\hypertarget{excelux5e94ux7528ux4e3eux4f8b}{}
\href{ch3Axn.xlsx}{\emph{利用生命表计算趸交纯保费示例}}
\end{block}
\end{frame}

\begin{frame}
借助\textbf{剩余寿命}函数计算\(\Ax{\termxn}\)

\begin{itemize}
\item
  (x)身故年末赔付1元的n年定期寿险
\item
  令K为(x)身故时的取整剩余寿命，显然K=0,1,2,\ldots,n-1
\item
  该定期寿险赔付金额现值Z是一个随机变量
\end{itemize}

\[
Z =  
\begin{cases}
v^{K+1}& K=0,1,2,\ldots,n-1 \\
0& K=n,n+1,...
\end{cases}
\]
\end{frame}

\begin{frame}
预备知识：

\begin{itemize}
\item
  随机变量X，其密度函数为\(f(x)\)
\item
  构造随机变量\(Y=\phi(X)\)，那么Y的期望\(E(Y)=\int{\phi(x)f(x)dx}\)
\item
  当X是离散分布时基本同理
\end{itemize}

简单应用：

\[
\begin{aligned}
\Ax{\termxn}&=E(Z)\\
&=\qx{x} v+\qx[1|]{x} v^2+ \ldots + \qx[n-1|]{x} v^n
\end{aligned}
\]
\end{frame}

\begin{frame}{即死即付定期寿险精算现值}
\protect\hypertarget{ux5373ux6b7bux5373ux4ed8ux5b9aux671fux5bffux9669ux7cbeux7b97ux73b0ux503c}{}
借助\textbf{剩余寿命}函数发现：

\begin{itemize}
\item
  (x)剩余寿命T为连续随机变量
\item
  (x)身故即得到1元保险金给付，其现值Z为
\end{itemize}

\[
Z=
\begin{cases}
v^T& 0\leq T \leq n \\
0& T>n 
\end{cases}
\]
\end{frame}

\begin{frame}
预备知识：

\begin{itemize}
\item
  随机变量X，其密度函数为\(f(x)\)
\item
  构造随机变量\(Y=\phi(X)\)，那么Y的期望\(E(Y)=\int{\phi(x)f(x)dx}\)
\item
  当X是离散分布时基本同理
\end{itemize}

简单应用：

\[
\begin{aligned}
\Ax*{\termxn}&=E(Z)\\
&=\int_0^n{v^t \px[t]{x}\mu_{x+t}dt}
\end{aligned}
\]
\end{frame}

\begin{frame}
不加证明地给出身故年末给付定期寿险和即死即付定期寿险的精算现值之间关系:

在身故人数均匀分布条件下

\[
\Ax*{\termxn}=\frac{i}{\delta}\Ax{\termxn}
\]

其中，\(i\)为年实际利率，\(\delta\)为\(i\)对应的利息力。
\end{frame}

\begin{frame}[fragile]{定期寿险例题}
\protect\hypertarget{ux5b9aux671fux5bffux9669ux4f8bux9898}{}
已知\(\lx{x}=100-x\)且\(0\leq x \leq 100\)，\(\delta=0.05\)，假设身故人数均匀分布。计算：(1)\(\Ax{\term{40}{25}}\)；(2)\(\Ax*{\term{40}{25}}\)

\pause

\textbf{最笨的办法}：按照题意生成一张生命表

\begin{Shaded}
\begin{Highlighting}[]
\NormalTok{x}\OtherTok{\textless{}{-}}\FunctionTok{c}\NormalTok{(}\DecValTok{0}\SpecialCharTok{:}\DecValTok{100}\NormalTok{)}\CommentTok{\#生成年龄}
\NormalTok{lx}\OtherTok{\textless{}{-}}\FunctionTok{c}\NormalTok{(}\DecValTok{100}\SpecialCharTok{:}\DecValTok{0}\NormalTok{)}\CommentTok{\#生成lx}
\NormalTok{lt}\OtherTok{\textless{}{-}}\FunctionTok{new}\NormalTok{(}\StringTok{"lifetable"}\NormalTok{,}\AttributeTok{x=}\NormalTok{x,}\AttributeTok{lx=}\NormalTok{lx)}\CommentTok{\#生成生命表}
\NormalTok{lt.act}\OtherTok{\textless{}{-}}\FunctionTok{new}\NormalTok{(}\StringTok{"actuarialtable"}\NormalTok{,}\AttributeTok{x=}\NormalTok{x,}\AttributeTok{lx=}\NormalTok{lx,}\AttributeTok{interest=}\FunctionTok{exp}\NormalTok{(}\FloatTok{0.05}\NormalTok{)}\SpecialCharTok{{-}}\DecValTok{1}\NormalTok{)}\CommentTok{\#生成精算表}
\FunctionTok{print}\NormalTok{(}\FunctionTok{paste}\NormalTok{(}\StringTok{"身故年末给付的定期寿险精算现值为"}\NormalTok{,}\FunctionTok{round}\NormalTok{(}\FunctionTok{Axn}\NormalTok{(lt.act,}\DecValTok{40}\NormalTok{,}\DecValTok{25}\NormalTok{),}\AttributeTok{digits=}\DecValTok{4}\NormalTok{),}\AttributeTok{sep=}\StringTok{""}\NormalTok{))}\CommentTok{\#计算身故年末给付的定期寿险精算现值}
\end{Highlighting}
\end{Shaded}

\begin{verbatim}
## [1] "身故年末给付的定期寿险精算现值为0.2319"
\end{verbatim}

\begin{Shaded}
\begin{Highlighting}[]
\FunctionTok{print}\NormalTok{(}\FunctionTok{paste}\NormalTok{(}\StringTok{"即死即付定期寿险精算现值为"}\NormalTok{,}\FunctionTok{round}\NormalTok{((}\FunctionTok{exp}\NormalTok{(}\FloatTok{0.05}\NormalTok{)}\SpecialCharTok{{-}}\DecValTok{1}\NormalTok{)}\SpecialCharTok{/}\FloatTok{0.05}\SpecialCharTok{*}\FunctionTok{Axn}\NormalTok{(lt.act,}\DecValTok{40}\NormalTok{,}\DecValTok{25}\NormalTok{),}\AttributeTok{digits=}\DecValTok{4}\NormalTok{),}\AttributeTok{sep=}\StringTok{""}\NormalTok{))}\CommentTok{\#身故人数均匀分布条件下，即死即付定期寿险精算现值}
\end{Highlighting}
\end{Shaded}

\begin{verbatim}
## [1] "即死即付定期寿险精算现值为0.2378"
\end{verbatim}
\end{frame}

\begin{frame}{定期寿险例题}
\protect\hypertarget{ux5b9aux671fux5bffux9669ux4f8bux9898-1}{}
已知\(\lx{x}=100-x\)且\(0\leq x \leq 100\)，\(\delta=0.05\)，假设身故人数均匀分布。计算：(1)\(\Ax{\term{40}{25}}\)；(2)\(\Ax*{\term{40}{25}}\)

\textbf{数学解析方法}：

利用公式\(\Ax*{\termxn}=\int_0^n{v^t \px[t]{x}\mu_{x+t}dt}\)

先求\(\px[t]{x}=\frac{100-x-t}{100-x}\)

再求\(\mu_{x+t}=-(\ln \px[t]{x})'=\frac{1}{100-x-t}\)

显然\(\px[t]{x}\mu_{x+t}=\frac{1}{100-x}\)，那么\(g(t|x=40)=1/60\)

\[
\Ax{\term{40}{25}}=\int_0^{25}{\frac{1}{60}\times e^{-0.05t}}dt=0.2318
\]
\end{frame}

\end{document}
